An object is placed in front of a concave mirror of radius of curvature 15 cm, at a distance of 10 cm, the position and nature of the image formed is :

+ 30 c m, v i r t u a l a n d e r e c t

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+ 30 c m, r e a l a n d i n v e r t e d

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- 30 c m, v i r t u a l a n d e r e c t

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- 30 c m, r e a l a n d i n v e r t e d

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Solution

The correct option is B $- 30 c m, r e a l a n d i n v e r t e d$
Given Radius of Curvature =15
$\begin{matrix} | u | = 10 c m \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \frac{- 2}{15} = \frac{1}{v} - \frac{1}{10} v = - 30 c m \end{matrix}$
$\begin{matrix} m = \frac{h_{i}}{h_{0}} \Rightarrow \frac{- v}{u} = \frac{h_{i}}{h_{0}} h_{P} = - \frac{(t - 3 ϕ)}{+ 1 \emptyset} h_{0} h_{Q} = - 3 h_{0} \end{matrix}$
So the image formed is inverted image.

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An object is placed in front of a concave mirror of radius of curvature 15 cm, at a distance of 10 cm, the position and nature of the image formed is :

The correct option is B −30cm,real and invertedGiven Radius of Curvature =15 |u|=10cm1f=1v+1u−215=1v−110v=−30 cmm=hih0⇒−vu=hih0hP=−(t−3ϕ)+1∅h0hQ=−3h0 So the image formed is inverted image.